In a wireless communication system, a base station can transmit/receive data to/from one or more subscribers simultaneously. Generally, a communication link from a base station to a subscriber is called as a downlink, and a communication link from a subscriber to a base station is called as an uplink. In the uplink, it is necessary for the subscriber to transmit uplink control signaling, such as a bandwidth request signal for requesting bandwidth, an acknowledgement signal for a downlink data package, etc. to the base station, so as to support data transmission in the downlink. In the system design, it is required to consider how to reduce the overhead of uplink control signaling transmitted on the uplink as much as possible so as to improve the spectral efficiency of the system.
The acknowledgement signal for the downlink data package is one kind of uplink control signaling, and is used in the communication between the base station and a mobile station. It has two states, i.e. ACK and NACK. As shown in FIG. 1, a data transmission process between the base station and the mobile station is illustrated. Firstly, the process starts from step S11, and the base station transmits data to the mobile station via a downlink at step S12. Then, the subscriber receives and decodes the data at step S13, and determines whether the decoding is correct at step S14. If the decoding is correct, the mobile station transmits to the base station a control signal of ACK indicating that the decoding is success at step S15. The base station may transmit a new data package to the mobile station at step S16, and the flow of this data transmission is terminated at step S19.
On the other hand, if it is determined at step S14 that the decoding is not correct, the mobile station transmits to the base station a control signal of NACK indicating that the data package decoding fails at step S17. The base station retransmits the previously transmitted data at step S18. Then the process returns to step S13 and the subscriber resumes to receive data.
As can be seen from the above description, the acknowledgement signal for the downlink data package is essentially a control signal which represents a successful data decoding (ACK) or a failed data decoding (NACK). For the purpose of easy explanation, the acknowledgement signal for the downlink data package hereafter may be abbreviated as a response signal or an ACK signal, as long as this does not incur any ambiguity.
When the fixed resources assigned by the base station are used to the transmission of uplink ACK signal, the ACK signal of multiple subscribers may be transmitted in multiplex through various approaches. The various approaches include, for example, Time Division Multiplex (TMD) in which the multiple subscribers occupy different time resources, Frequency Division Multiplex (FDM) in which the multiple subscribers occupy different frequency resources and Code Division Multiplex (CDM) in which the multiple subscribers occupy different code word resources. In comparison with TDM and FDM, to employ the resource assignment approach of CDM can provide assignable resources more effectively and provide a better performance in bit error rate of uplink control signals, and thus are applied widely in today's wireless communication systems. For example, the Long Term Evolution (LTE) system of 3GPP has been specified to take CDM as a frequency-domain multiplex approach for an uplink ACK channel.
To transmit as more uplink control signals as possible on the assigned fixed resources, LTE also employs a time-domain spread spectrum approach. In the time-domain spread spectrum approach, an ACK bit of each subscriber is represented uniquely with a frequency-domain sequence and a time-domain sequence, and ACK bits of multiple subscribers can be multiplexed on the same resources with different frequency-domain and/or time-domain sequences and then be transmitted simultaneously.
FIG. 2 illustrates a typical multiplex approach for an uplink ACK sequence. In FIG. 2, each sub-frame includes seven Orthogonal Frequency Division Multiplexing (OFDM) symbols. The intermediate three OFDM symbols are used for transmitting pilot signals of multiple subscribers, and the remaining four symbols are used for transmitting ACK signal of the multiple subscribers.
A Constant Amplitude Zero Auto Correlation (CAZAC) sequence may be used as a frequency-domain sequence for an ACK signal or a pilot signal. The CAZAC sequence is characterized in the followings: 1) constant amplitude, that is, each element of each CAZAC sequence has the same amplitude value; and 2) zero cyclic auto-correlation, that is, the CAZAC sequences with different cyclic time shifts has zero correlation. The CAZAC sequence includes sequences such as Zadoff-Chu (ZC) sequence and General Chirp-like (GCL) sequence, etc., of which the respective sequence description may be obtained by reference to a Non-Patent Document “Ghosh. A, Ratasuk. R, Uplink control channel design for 3GPP LTE. IEEE PIMRC 2007”.
A ZC sequence is described here for exemplary purpose. An expression of the ZC sequence may be represented as:
      p    ⁡          (      n      )        =      {                                        ⅇ                                                                                                                              -                        j                                            ⁢                                                                        2                          ⁢                          π                          ⁢                                                                                                          ⁢                          l                                                M                                            ⁢                                              (                                                  n                          +                                                      n                            ⁢                                                                                          n                                +                                1                                                            2                                                                                                      )                                                                                                                                                if                        ⁢                                                                                                  ⁢                        M                        ⁢                                                                                                                                  ⁢                                                                                                                                ⁢                        is                        ⁢                                                                                                                                  ⁢                                                                                                                                ⁢                        odd                                            ,                                                                                          ⁢                                              n                        =                        0                                            ,                      1                      ,                                                                                          ⁢                      …                      ⁢                                                                                          ,                                              M                        -                        1                                                                                                        ⁢                                                                                                                    ⅇ                                                                                                      -                      j                                        ⁢                                                                  2                        ⁢                        π                        ⁢                                                                                                  ⁢                        l                                            M                                        ⁢                                          (                                              n                        +                                                                              n                            2                                                    2                                                                    )                                                                                                                                  if                      ⁢                                                                                          ⁢                      M                      ⁢                                                                                                                        ⁢                                                                                                                      ⁢                      is                      ⁢                                                                                          ⁢                      even                                        ,                                                                                  ⁢                                          n                      =                      0                                        ,                    1                    ,                                                                                  ⁢                    …                    ⁢                                                                                  ,                                          M                      -                      1                                                                                                              where M is a sequence length, l is an arbitrary integer, and M and l are prime numbers with respect to each other.
In FIG. 2, F−1 denotes an IFFT operation, and the time-domain sequence for the ACK bits are denoted as w0-w3. A vector in a Hadamard matrix may be used as the time-domain sequence. A 4×4 Hadamard matrix may be represented as
      [                            1                          1                          1                          1                                      1                                      -            1                                    1                                      -            1                                                1                          1                                      -            1                                                -            1                                                1                                      -            1                                                -            1                                    1                      ]    .The value of w0-w3 is a row vector of the Hadamard matrix. That is, if the first row vector is selected as the time-domain sequence, w0-w3=[1,1,1,1]. The four row vectors of the Hadamard matrix are assigned index numbers respectively, that is, time-domain sequence 1 corresponds to a sequence [1,1,1,1], time-domain sequence 2 corresponds to a sequence [1,−1,1,−1], and so on.
The resources used for ACK signal may be continuous or non-continuous sub-carriers in the frequency domain. For example, the continuous sub-carriers are assigned as the resources, and it is assumed that these resources locate at the boundaries of the Orthogonal Frequency Division Multiplexing (OFDM) symbols as shown in FIG. 3. If twelve continuous sub-carriers are used in the frequency domain, there may be in theory twelve ZC sequences with different time shifts as frequency-domain sequences for transmitting the ACK bits. However, in practice, the ZC sequences are generally taken every two time shifts among all the ZC sequences to reduce interference, so that there are six available frequency-domain sequences.
In the time domain, three OFDM symbols are used as pilot symbols, and the remaining four symbols are used as data symbols. The pilot signals are generally used for estimating each subscriber's channel. Similar with the ACK signal, the pilot signal of each subscriber also corresponds to a frequency-domain sequence and a time-domain sequence.
In the view of pilot, there are three time-domain sequences and thus the number of the ACK bits that can be transmitted simultaneously is 6*3=18. In the view of data, since the number of the occupied time-domain OFDM symbols is 4, and six ZC sequences with different cyclic time shifts are used as the frequency-domain sequences, the number of 1-bit ACK signal that can be transmitted simultaneously is 6*4=24. Due to the one-to-one corresponding relationship between the pilots and the ACK data, the number of the ACK signal that can be transmitted simultaneously can only be the minimum value, i.e. eighteen. A resource assignment solution can be represented by table 1.
TABLE 1ACK bit sequence assignmentindex of time-pilot/datadomain sequenceZC sequences with differentCyclic time shifts1234117132328144539156741016895111710116121812
In table 1, the index of the time-domain and frequency-domain sequences assigned to subscriber 1 is (1,1), and the index of the time-domain and frequency-domain sequences assigned to subscriber 2 is (1,3). Totally eighteen subscribers can transmit 1 bit of ACK signal simultaneously The ACK bit transmitted by each subscriber occupies a different time-domain sequence and/or a different frequency-domain sequence.
After receiving the ACK signal simultaneously transmitted by the multiple subscribers, the base station employs a typical receiving method as shown in FIG. 4. The receiving process of the base station is described as follows by an example in which the base station receives the ACK signal of a certain subscriber k. As shown in FIG. 4, after receiving the multiplexed ACK data of the multiple subscribers, the base station multiplies each of the received OFDM symbols by the time-domain sequence (w0,w1,w2,w3) of the subscriber k and sums the resulting sequences at step S41, and then executes an FFT operation on the resulting sequence at step S42. Then, at step S43, the base station performs an equalization on the resulting sequence according to channel estimation value obtained in advance from the pilots of the subscriber to eliminate the influence of a wireless channel, and then performs a correlating process on the resulting sequence with a frequency-domain sequence p(n) corresponding to the subscriber k at step S44. Finally, the base station compares at step S45 the result obtained at step S44 with a predefined threshold value. If the result is greater than the threshold value, it means that an NACK signal is transmitted by the subscriber k. If the result is less than the threshold value, it means that an ACK signal is transmitted by the subscriber k.
The case as described above is a case that each subscriber feeds back only one ACK signal, i.e. one ACK bit. However, in many cases, some subscribers in the system need to feed back multiple bits of ACK signal simultaneously, such as in a Multi-Codewords (MCW) multiple input multiple output (MIMO) system or a Time Division Duplexing (TDD) system, etc. In the case that some subscribers need to feed back multiple bits of ACK signal simultaneously, how to flexibly assign the time-domain sequence and the frequency-domain sequence corresponding an uplink ACK bit in the system has become an interesting focus.